How do you find all the zeros of #f(x)=x^3-3x^2+x-3#?

1 Answer
Nov 2, 2016

Answer:

#f(x)# has zeros #x=3# and #x=+-i#

Explanation:

The difference of squares identity can be written:

#a^2-b^2=(a-b)(a+b)#

We use this later with #a=x# and #b=i#

#f(x) = x^3-3x^2+x-3#

Note that the ratio of the first and second terms is the same as that of the third and fourth terms, so this cubic will factor by grouping:

#x^3-3x^2+x-3 = (x^3-3x^2)+(x-3)#

#color(white)(x^3-3x^2+x-3) = x^2(x-3)+1(x-3)#

#color(white)(x^3-3x^2+x-3) = (x^2+1)(x-3)#

#color(white)(x^3-3x^2+x-3) = (x^2-i^2)(x-3)#

#color(white)(x^3-3x^2+x-3) = (x-i)(x+i)(x-3)#

Hence zeros:

#x = +-i#

#x = 3#